Problem 129 - Advanced Problems in Organic Chemistry for Competitive Examinations

Question

An organic compound \(\mathrm{A}\left(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Cl}\right)\) is optically active and on treatment with ethanolic KOH solution yields \(\mathrm{B}\left(\mathrm{C}_{5} \mathrm{H}_{10}\right)\) as a major product, which does not show stereoisomerism. Also A on treatment with \(\left(\mathrm{CH}_{3}\right)_{2}\) CuLi yields \(\mathrm{C}\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\), which is optically inactive. Deduce structures of \(\mathrm{A}\) to \(\mathrm{C}\). Identify structure of "B" compound

Step-by-Step Solution

Answer

B is pent-1-ene.

Step 1: Determine possible structure of compound A

The compound A (C\(_5\)H\(_{11}\)Cl) is optically active, meaning it has a chiral center. The presence of chlorine suggests it could be a haloalkane. Given the molecular formula, A could be 2-chloropentane or 3-chloropentane, but since A is optically active, it must have a chiral center. This means 2-chloropentane is a suitable candidate because the second carbon is chiral.

Step 2: Analyze the reaction with ethanolic KOH

When A (2-chloropentane) is treated with ethanolic KOH, it undergoes an elimination reaction to form B. Ethanolic KOH tends to promote elimination reactions via the E2 mechanism, resulting in the formation of an alkene (C\(_5\)H\(_{10}\)). Since B is the major product and does not show stereoisomerism, this indicates the formation of pentene without stereoisomers, which suggests B is pent-1-ene, as it would have no E/Z isomerism.

Step 3: Evaluate the reaction with (CH3)2CuLi

Compound A reacts with (CH\(_3\))\(_2\)CuLi, a reagent known to perform substitution reactions replacing the halogen with an alkyl group, and forms C (C\(_6\)H\(_{14}\)). Compound C is saturated (alkane) and optically inactive, which suggests the removal of the chiral center in this reaction. Given that C shows no optical activity, it confirms the substitution left a non-chiral product, which is n-hexane.

Key Concepts

Optical Activity

Optical activity is a unique property of certain organic compounds. It implies that the molecule can rotate plane-polarized light. This occurs due to the presence of a chiral center within the molecule. A chiral center is a carbon atom bonded to four different groups, leading to non-superimposable mirror images, also called enantiomers. Each enantiomer can rotate light in a different direction: clockwise for dextrorotatory (+) or counterclockwise for levorotatory (-).
Consider compound A (C sub{5}H sub{11}Cl) in our exercise; it is optically active because of its chiral center. This inherent chiral property differentiates its potential structures into specific configurations, such as 2-chloropentane. The chirality of a molecule is crucial in determining its optical activity and subsequently the behaviors and reactions it can participate in.

Elimination Reaction

Elimination reactions play a central role in organic synthesis. These reactions involve the removal of atoms or groups from a molecule, forming a double bond, typically resulting in an alkene. For our compound A, when treated with ethanolic KOH, it undergoes an elimination reaction.
Ethanolic KOH promotes an E2 elimination mechanism. During this process, both the leaving group (like chlorine in 2-chloropentane) and a hydrogen atom from an adjacent carbon are removed simultaneously, forming a double bond. This results in the creation of compound B (C sub{5}H sub{10}). Specifically, B is identified as pent-1-ene since it has no stereoisomerism, indicating a straight-chain alkene without substituents that can lead to E/Z isomerism.
Elimination reactions are fundamental in forming unsaturated compounds and are a critical step when transitioning from saturated to unsaturated organic molecules.

Nucleophilic Substitution

Nucleophilic substitution is another important type of reaction in organic chemistry, where one atom or group in a molecule is replaced by another. A common class of these reactions involves the replacement of a halogen atom by a nucleophile.
In the given exercise, compound A reacts with (CH sub{3}) sub{2}CuLi, which is known for undergoing nucleophilic substitution. Here, the dimethylcopper lithium reagent replaces the chlorine atom in 2-chloropentane with a methyl group. This forms compound C (C sub{6}H sub{14}), specifically n-hexane, since all parts of the alkane are fully saturated and lack a chiral center.
The result is a product that is optically inactive, as the symmetrical nature of n-hexane means no chiral centers remain. Understanding nucleophilic substitution helps in synthesizing a variety of compounds by strategically altering molecular structures.

Chiral Center

The concept of a chiral center is crucial for understanding optical activity in organic compounds. A chiral center is typically a tetrahedral carbon atom bonded to four distinct groups. This arrangement allows for two non-identical mirror image forms or enantiomers.
In the exercise, compound A is optically active because it features a chiral center. Taking 2-chloropentane as an example, the second carbon serves as the chiral center. Having one chiral center means the molecule can exist in two enantiomeric forms, leading to its unique property of optical activity.
Recognizing and identifying chiral centers helps chemists design and predict the behavior of organic compounds, particularly in producing specific isomers that can react or interact distinctly in biological and chemical systems. Knowing about chiral centers can also enhance understanding of molecular interactions in pharmaceuticals, fragrances, and other applications.